Problem 318 Solve. \((5 b+1)(6 b+1)=0\)... [FREE SOLUTION]

Chapter 7: Problem 318

Solve. \((5 b+1)(6 b+1)=0\)

Short Answer

Expert verified

\(b = -\frac{1}{5}\) or \(b = -\frac{1}{6}\)

Step by step solution

Apply the Zero Product Property

The zero product property states that if the product of two numbers is zero, then at least one of the numbers must be zero. Set each factor in the equation \((5b + 1)(6b + 1) = 0\) to zero. This gives us two equations:1. \(5b + 1 = 0\)2. \(6b + 1 = 0\)

Solve for b in the first equation

Solve the first equation:\(5b + 1 = 0\).Subtract 1 from both sides of the equation:\(5b = -1\).Now, divide both sides by 5:\(b = -\frac{1}{5}\).

Solve for b in the second equation

Solve the second equation:\(6b + 1 = 0\).Subtract 1 from both sides of the equation:\(6b = -1\).Now, divide both sides by 6:\(b = -\frac{1}{6}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Equations

Factoring is an essential skill in algebra. It involves breaking down complex expressions into simpler factors. For example, consider the equation (5b + 1)(6b + 1) = 0.Here, (5b + 1) and (6b + 1) are factors.
The aim is to write the equation in a form where each factor could be zero. Factoring makes equations easier to solve.
It's important to understand that if two numbers multiply to give zero, one or both of them must be zero. This is a powerful property used in algebra, called the Zero Product Property.

Solving Linear Equations

Solving linear equations involves finding the value of the unknown variable that makes the equation true.
Let's look at the equation from the exercise: 5b + 1 = 0.
To solve it, follow these steps:

Isolate the variable b
Subtract 1 from both sides to get: 5b = -1.
Divide both sides by 5: b = -1/5.

Similarly, solve the second equation: 6b + 1 = 0.
Again, isolate b: 6b = -1. Divide both sides by 6 to get: b = -1/6.
These steps show the systematic process of solving linear equations by isolating the variable.

Algebraic Solutions

Algebraic solutions involve solving equations using algebraic methods.
These methods require understanding and applying algebraic properties and operations.
In the exercise, we used the Zero Product Property and basic operations like addition, subtraction, and division.
Steps to solve algebraically:

Identify and set each factor to zero.
Solve for the unknown variable in each individual equation.

This approach can be applied to various types of algebraic equations. It's a reliable method to find solutions systematically and accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Solve. \((5 b+1)(6 b+1)=0\)

Short Answer

Step by step solution

Apply the Zero Product Property

Solve for b in the first equation

Solve for b in the second equation

Key Concepts

Factoring Equations

Solving Linear Equations

Algebraic Solutions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Geometry

Applied Mathematics

Pure Maths

Decision Maths

Study anywhere. Anytime. Across all devices.